Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A driver moving at 36 km/h sees an obstacle. Her reaction time is 0.5 s, after which the car brakes with a constant deceleration of 6 m/s². Find the total stopping distance d from the moment she sees the obstacle.
A stone is dropped from rest into a well. The splash is heard 1.7 s after the stone is released. The speed of sound is 343 m/s; ignore air resistance and use g = 9.81 m/s². Find the depth d of the well.
Masses of 8.5 kg and 8 kg hang on either side of a light, frictionless pulley, connected by a light string. The system is released from rest. Use g = 9.81 m/s². Find the tension T in the string.
A 4 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 2 kg mass. The coefficient of kinetic friction between block and table is 0.3. Use g = 9.81 m/s². Find the tension T in the string.
A ball is launched at 35° above horizontal from level ground. Ignore air resistance. Find the maximum height H the ball reaches.
A ball is launched at 45° above horizontal from the top of a 5 m cliff. Ignore air resistance. Find the horizontal distance x travelled before the ball hits the ground.
A small ball on a string of length 1.1 m moves in a horizontal circle so that the string makes a constant angle of 15° with the vertical (a conical pendulum). Use g = 9.81 m/s². Find the speed v of the ball.
A curve of radius 50 m is banked at 30°, and the coefficient of static friction between tyres and road is 0.4. Use g = 9.81 m/s². Find the lowest speed vmin at which a car can take the curve without sliding down the bank.
A block slides from rest down a slope at 30° to the horizontal. The coefficient of kinetic friction is 0.35. Find its speed v after it has slid 20 m along the slope.
A 5 kg block is released from rest at height 2 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 4 m long (μ = 0.3) and then runs onto a spring (k = 5000 N/m) on a smooth surface. Find the maximum compression x of the spring.
A 15 g bullet is fired horizontally into a 4 kg wooden block hanging on light strings, and stays inside it. The block swings up until it is 22 cm higher than at the start. Use g = 9.81 m/s². Find the speed v of the bullet before impact.
At a crossing a 1900 kg car travelling east at 20 m/s collides with a 1900 kg car travelling north at 9 m/s. The cars lock together. Find the speed v′ of the wreck just after the collision.
A flywheel with a moment of inertia of 24 kg·m² spins at 500 rpm. A brake applies a constant friction torque of 155 N·m. How long t does it take to stop?
A person stands on a frictionless turntable and holds a 3 kg dumbbell in each hand, 0.65 m from the axis. The person and turntable alone have a moment of inertia of 1.5 kg·m² about the axis. They turn at 35 rpm, then pull the dumbbells in to 0.2 m from the axis. Treat the dumbbells as point masses. Find the new speed n2 in rpm.
A belt runs on a 100 mm driving pulley at 1500 rpm. The tension is 600 N on the tight side and 180 N on the slack side. Find the power P transmitted, in kW.
An open belt connects a 125 mm driving pulley at 960 rpm to a 250 mm pulley; the centres are 600 mm apart. The coefficient of friction is 0.3 and the tight-side tension may not exceed 800 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).
How long must a simple pendulum be to have a period of 2.4 s? Use g = 9.81 m/s². Find L.
A 15 g bullet moving at 350 m/s hits and stays in a 0.5 kg block resting on a frictionless surface. The block is attached to a horizontal spring with k = 800 N/m. Find the amplitude A of the resulting oscillation in cm.
In a free-vibration test of a damped mass–spring system, a displacement peak of 25 mm is followed 3 cycles later by a peak of 17.1 mm. Find the damping ratio ζ.
A 50 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 5 kN/m and the lower one k₂ = 50 kN/m. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
Answer: d = 13.33 m
Motion with constant acceleration: v = v₀ + a·t, s = v₀·t + ½·a·t² and v² = v₀² + 2·a·s. Pick the equation that contains the three things you know and the one you want, and keep the signs consistent with one positive direction.
Answer: d = 13.53 m
Motion with constant acceleration: v = v₀ + a·t, s = v₀·t + ½·a·t² and v² = v₀² + 2·a·s. Pick the equation that contains the three things you know and the one you want, and keep the signs consistent with one positive direction.
Only the positive root is physical.
Answer: T = 80.86 N
Newton’s second law: the net force on a body equals its mass times its acceleration, ΣF = m·a. Draw a free-body diagram, choose a positive direction along the motion, and add every force in that direction: applied forces, friction μ·N against the motion, and the components of gravity.
The net driving force is the difference of the weights; it moves the total mass.
Answer: T = 17 N
Newton’s second law: the net force on a body equals its mass times its acceleration, ΣF = m·a. Draw a free-body diagram, choose a positive direction along the motion, and add every force in that direction: applied forces, friction μ·N against the motion, and the components of gravity.
Answer: H = 1.677 m
Projectile motion: with no air resistance, x = vxt and y = h + vyt − ½gt². The two directions are independent except that they share the same time.
Horizontal velocity stays constant; vertical motion has a constant acceleration g = 9.81 m/s² downward.
At the top the vertical velocity is zero: vy² = 2gH.
Answer: x = 30.39 m
Projectile motion: with no air resistance, x = vxt and y = h + vyt − ½gt². The two directions are independent except that they share the same time.
Horizontal velocity stays constant; vertical motion has a constant acceleration g = 9.81 m/s² downward.
Solve y(t) = h + vy·t − ½gt² = 0 for the positive root.
Answer: v = 0.8651 m/s
Uniform circular motion: an object moving at speed v on a circle of radius r accelerates towards the centre with a = v²/r, so it needs a net inward force F = m·v²/r. That force is supplied by something real: friction, tension, gravity or a normal force.
Tension and gravity together give the inward force: tan θ = v²/(r·g).
Answer: vmin = 8.407 m/s
Uniform circular motion: an object moving at speed v on a circle of radius r accelerates towards the centre with a = v²/r, so it needs a net inward force F = m·v²/r. That force is supplied by something real: friction, tension, gravity or a normal force.
Friction points up the slope.
Answer: v = 8.79 m/s
Work–energy principle: the change in kinetic energy equals the work done by all forces. Writing gravity and springs as potential energy (mgh and ½kx²) turns this into a balance: initial energy = final energy + energy lost to friction (μN·d).
The block drops a height d·sin θ while friction removes μmg·cos θ·d. The mass cancels.
Answer: x = 12.53 cm
Work–energy principle: the change in kinetic energy equals the work done by all forces. Writing gravity and springs as potential energy (mgh and ½kx²) turns this into a balance: initial energy = final energy + energy lost to friction (μN·d).
The ramp is smooth, so only the strip removes energy.
Answer: v = 556.1 m/s
Conservation of momentum: when no external force acts during a collision, the total momentum Σm·v before equals the total after. Velocities have a direction, so pick a positive direction and give velocities the other way a minus sign. In an elastic collision kinetic energy is conserved too; in a perfectly inelastic one the bodies stick together.
Answer: v′ = 10.97 m/s
Conservation of momentum: when no external force acts during a collision, the total momentum Σm·v before equals the total after. Velocities have a direction, so pick a positive direction and give velocities the other way a minus sign. In an elastic collision kinetic energy is conserved too; in a perfectly inelastic one the bodies stick together.
Answer: t = 8.107 s
Rotation about a fixed axis: the sum of the moments about the axis equals I·α (ΣM = I·α), just as ΣF = m·a for translation. In energy form, a torque T turning through an angle θ (in radians) does work T·θ, which changes the kinetic energy ½·I·ω²; gravity adds or removes m·g·Δh of the centre of mass. Power is P = T·ω.
Answer: n2 = 81.16 rpm
Rolling and angular momentum: a body rolling without slipping has v = ω·r, so its kinetic energy is ½·m·v² + ½·I·ω² = ½·m·v²·(1 + I/(m·r²)). Down a slope that gives a = g·sin θ/(1 + I/(m·r²)). When no external moment acts about an axis, the angular momentum about that axis stays constant, I₁·ω₁ = I₂·ω₂, even if kinetic energy is lost.
The kinetic energy increases: the extra energy comes from the work the person does pulling the dumbbells in.
Answer: P = 3.299 kW
Belt drives: without slip, both pulleys have the same rim speed v = π·d·n/60, so n₁·d₁ = n₂·d₂ and the speed ratio is d₂/d₁. Slip s makes the driven pulley a little slower: n₂ = n₁·(d₁/d₂)·(1 − s). The power transmitted is P = (T₁ − T₂)·v, with T₁ the tension on the tight side and T₂ on the slack side. Friction limits the ratio of the tensions: T₁/T₂ = eμθ, where θ is the angle of wrap in radians on the smaller pulley; for an open belt θ = π − 2·asin((d₂ − d₁)/(2C)) with centre distance C.
Answer: P = 2.941 kW
Belt drives: without slip, both pulleys have the same rim speed v = π·d·n/60, so n₁·d₁ = n₂·d₂ and the speed ratio is d₂/d₁. Slip s makes the driven pulley a little slower: n₂ = n₁·(d₁/d₂)·(1 − s). The power transmitted is P = (T₁ − T₂)·v, with T₁ the tension on the tight side and T₂ on the slack side. Friction limits the ratio of the tensions: T₁/T₂ = eμθ, where θ is the angle of wrap in radians on the smaller pulley; for an open belt θ = π − 2·asin((d₂ − d₁)/(2C)) with centre distance C.
Answer: L = 1.431 m
Simple harmonic motion: a mass m on a spring of stiffness k oscillates with angular frequency ω = √(k/m) and period T = 2π/ω = 2π·√(m/k). A simple pendulum of length L has T = 2π·√(L/g). With amplitude A, the largest speed is A·ω, the largest acceleration A·ω² and the energy ½·k·A².
Answer: A = 25.86 cm
Simple harmonic motion: a mass m on a spring of stiffness k oscillates with angular frequency ω = √(k/m) and period T = 2π/ω = 2π·√(m/k). A simple pendulum of length L has T = 2π·√(L/g). With amplitude A, the largest speed is A·ω, the largest acceleration A·ω² and the energy ½·k·A².
Answer: ζ = 0.02014
Free vibration of a mass–spring system: the undamped natural frequency is ωn = √(k/m) in rad/s, and fn = ωn/(2π) in Hz. A viscous damper c gives the damping ratio ζ = c/cc, with critical damping cc = 2·√(k·m) = 2·m·ωn. An underdamped system (ζ < 1) oscillates at ωd = ωn·√(1 − ζ²). Springs in parallel add, k = k₁ + k₂; springs in series add as reciprocals, 1/k = 1/k₁ + 1/k₂.
Answer: fn = 1.517 Hz
Free vibration of a mass–spring system: the undamped natural frequency is ωn = √(k/m) in rad/s, and fn = ωn/(2π) in Hz. A viscous damper c gives the damping ratio ζ = c/cc, with critical damping cc = 2·√(k·m) = 2·m·ωn. An underdamped system (ζ < 1) oscillates at ωd = ωn·√(1 − ζ²). Springs in parallel add, k = k₁ + k₂; springs in series add as reciprocals, 1/k = 1/k₁ + 1/k₂.
Springs in series share the same force and add their deflections, so they combine as reciprocals: 1/k = 1/k1 + 1/k2.
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